text:
Can an electric aircraft self-recharge?
I'm trying to design an aircraft with a long range, and I thought that an electric aircraft with a generator powered by the motor would be able to do that.
However, I've seen very few of these commercially not seen any aircraft that do this, and I was wondering why.
Is a fully electric aircraft too expensive/ not powerful enough to be feasible, economically or physically?
Also, is there possibly a way to generate enough power? An alternator/Magneto combination, perhaps?

What kind of aircraft are your asking about? We have a question about large electric aircraft feasibility, and for replacing jet engines.

Your question is too broad. Electric aircraft exist at varying scales, but without a more focused question it's hard to say if it's "feasible".

I mean a fairly small aircraft. Not much larger than a  Cessna. It would use an Electric Ducted Fan design, so it almost would be replacing a jet, but still using a prop.

@Nicholas: EDF is super inefficient. You will get at minimum 4x better endurance and at least 2x better range switching to props. The exception is short/open ducting like those found on paraglider backpacks.

Until you understand simple things (e.g., you can't build a perpetual motion machine), you won't get very far with your attempt to do a complicated thing (design a long-range aircraft).

Related: http://aviation.stackexchange.com/q/25933/4108

Could you be more specific and add more detail?  Right now it's unclear what you mean by "an electric aircraft with a generator powered by the motor", which to me sounds like an attempt to have an electric power source power a motor, with the motor driving a generator to get free energy out of the system.   Your question about "is there a way to generate enough power" is also unclear, since it sounds like you want to generate more electric power while on the plane, which would be a strange thing to do, mostly due to the difficulty in making it worth the weight.

This is really a physics question, not an aviation one: You're asking if you can build a perpetual motion machine (an aircraft that can indefinitely power itself), and the answer is no: A generator powered by the motor would require the motor to produce more power (to turn the generator in addition to the normal engine function), and combined with mechanical losses would ultimately reduce range because you're wasting power turning the generator. There is no such thing as free energy.

There is no such thing as free energy/perpetual motion so such a system is not possible. If you want a system to keep running you have to put energy into it.
As ymb1 pointed out, there's this thing called the conservation of energy and the laws of physics are pretty strict about it.
All systems lose energy as they perform work and that energy needs to be replaced from external resources.
It is possible to power a long range aircraft entirely from electricity
What you may be interested in is the solar impulse project which is an all electric (light) aircraft that is powered entirely by electric generated by solar panels on the wings.
It was the first solar-powered aircraft to make a round-the-world flight.

In my 8th grade physics class long ago the instructor told us if we were to take a small toy motor and spin its shaft by some mechanism it will produce small electric current, which obviously is the other way around usually.
I asked him if i could take one motor and power it with a battery then couple that motor with another motor will that provide enough current after some time to then power the first motor without battery? He said try it in your next lab.
I sure did, and the results? As soon as I disconnected the battery the first motor always went off even though in my (erroneous ) theory the other motor which was spinning at the same speed should have been sending some current to keep the system working.
So no, if you cant really power such a tiny milliamps motor like that forget about a motor which will power a big propeller for sustained flight.
My R/C aircrafts drain amperes like anything and the quicker you want to recharge them the more power your charger requires and that certainly can not come from those motors themselves.
Yes you can design your aircraft to be able to self charge to increase its flight duration up-to a certain extent and a number of electric aircrafts do this already but that can be a little extension and not self sustained for any considerable duration. KERS does that, Ram Air turbines do that. Even gliders have dynamo to that effect but they don't provide as much power as would be required by your proposed idea

You have a source of energy (the battery), you convert electrical energy into mechanical energy (first motor rotation), you transfer this mechanical energy to the second motor (dynamo) by coupling rotation axis. The mechanical energy of the second motor is converted back into electricity. But nothing is 100% efficient and there is an energy "loss" (actually heat generation) at each stage (conversion or transfer), so the final electrical energy is not sufficient to run the first motor at the same speed, so in a few seconds all levels of energy decrease to zero in a reaction loop.

And that's exactly my point - Such a system is not possible with current knowledge with reasonable number of resources

Ah, sorry then. "When" there would be no loss, the physicians' Grail would have been found but this would create a big trouble.


text:
How long does downwash last?
I think most of us know how lift is created. Air is deflected downward because of the shape of an airfoil.
How long does downwash last, is the air going to flow downward forever or will it climb back up to leveled degree? If so how long would it take for the air to get back on its normal flow path, which is not downward?

Related: How complete is our understanding of lift?

Like everything else, air will be accelerated either upwards or downwards if a force is applied to it. The aircraft's wing applies a downward force to the air mass thus accelerating it in that direction. However, nothing applies an upward force to that mass of air, so it can be assumed it will not climb back up.
(I am neglecting the formation of vortexes and possible differences in air density)
The velocity of the mass of air will be reduced by friction with and the inertia of the surrounding air until it comes to rest.
What you call the "normal flow path" of the air is it simply being still before an aircraft flies through it.

The forces that slow the air movement are proportional to the velocity of the movement, so the vortex decays slower the smaller it is and never decays completely. But of course after some time it decays to slower than turbulence caused by other reasons and then we can say it no longer exists for any practical purpose.
I don't know how quickly exactly the vortex decays. However, rough idea can be obtained from the separation requirements for aircraft.
The required separation for heavy and super behind same class is 4 NM, which for those aircraft is usually 1 to 1½ minute. After that time, the vortex has decayed enough to allow aircraft of similar weight as the one that made it, but not smaller ones.
For the biggest difference, small behind super, 8 NM is needed. Small aircraft are usually also slower, so say 3 to 4 minutes. After that time, the vortex has decayed so that it is no longer dangerous even to small aircraft, but probably still noticeable.
Now super is 30 times heavier than the largest small (A380 is 575t, small is up to 19t) and the vortex has to decay on similar order for the number to make sense. Probably even more, because the quantity that matter is moment of inertia and that grows faster than weight.
Since exponential means it decays by the same fraction every n minutes, after another couple of minutes it will be another two orders of magnitude down and likely no longer measurable.
So my wild guess is from under a minute behind small aircraft to maybe 10-15 minutes behind A380—and depending on speed; the vortex is weaker when flying fast.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What happens if a commercial airplanes navigation system completely failed during a flight?
What would happen if all of the the navigation systems (such as GPS & INS) in a commercial aircraft completely failed?
Could the aircraft land safely without all the navigation aids?

Aircraft have multiple navigation systems available, based on different technologies, power sources etc. It might be good if you can give a more specific scenario.

I wasn't sure if you're asking a technical question or as a concerned airline passenger.  Either way, you could look out the window and look around.

There are multiple navigation systems on a commercial aircraft the full scenario is covered here What will happen if satellite navigation fails?

The plane might land safely in a totally different airport than expected. Turkish Airlines Flight 1123, Aug 14th 2008

Hi.  Welcome to Aviation.SE.  Sorry for the downvotes, personally, I can't see why.  As to the close votes, I don't agree and think it's a perfectly reasonable question from someone (and I do make an assumption here) who does not necessarily understand modern aircraft systems.

@Simon: The problem I see is the lack of definition of "navigation system", e.g. does that include the failure of the compass? So not off-topic, but unclear for the moment.

If we want this site to be accessible to non-experts, we cannot expect people to come in and detail every individual navigation component - surely just saying "the navigation system" is enough to understand what he is asking... regardless though I feel this is enough of a duplicate to the linked question to be closed.

@mins If we edit the title to say all electric navigation systems would that help?

@Notts90: What is the OP after? They want to know if an airliner is able to land safely at an appropriate airport if automated navigation fails, perhaps including approach systems (VOR, ILS), and indeed it could, with vectors and PAR from ATC, even without knowing where the north is. But the radio needs to work. If not it could be helped be a side military aircraft and visual gestures. So I don't really know which case is to be explained, and whether this is at night or at day. Do you?

@mins I think it's basically what you just said, if all electric navigation aids failed (therefore also any automation would not work) would the plane land safely? As you say, the answer is probably. The idea of a military aircraft assisting is good too.

There was a film about a similar scenario in a private aircraft over the ocean. They came up with some interesting (though soundly biased) ideas for working out current location. http://m.imdb.com/title/tt0107556/plotsummary?ref_=m_tt_ov_pl

@Notts90: Perhaps. I'd wait a few hours to give the possibility for the OP to edit the question, and after would assume something and put it in the post. Just, it must be very clear, so that answers are using the same scenario. Or maybe it could be answered by describing solutions from the most basic (follow me aircraft) to ones that need some means aboard the plane.

I'm nominating this question for reopening. As of its current state, it is clear, on topic, and totally reasonable for everybody outside the aviation community to ask.

I would assume that commercial airlines would be able to get directions from Air Traffic Control.

Option 1: Use VFR to fly after contacting ATC, Option 2: Mayday -

The possible answers to your question are for all practical purposes infinite depending on the details of the system failure, the weather, where you are, and crew experience, to name a few. However, as a thought exercise extremely unlikely to ever happen in reality, let's play with it a bit.
First, let's define the navigational systems failure to be such that we cannot determine our position by reference to instruments in the cockpit. What have we got left? If we can still communicate, others can work with us to determine our position and develop a course of action.
The simplest case would be if we were talking to ATC, were in an area of secondary radar coverage, and our transponder was still working. If that were the case, it would be a matter of getting radar vectors to wherever we wanted to go. If I were the captain, I would want to go to the nearest suitable airport since if I had had a failure of everything that could tell me in-cockpit where I was, I would be worried about possible further failures.
If talking to ATC and in an area of primary radar coverage only or you also had transponder failure, they could give us a series of turns to identify us (used to do that back in the 1960s occasionally).
If you were out of range of VHF communications with ATC, you could try getting a relay via another aircraft that was in VHF range of ATC. Two aircraft at altitude can talk to each other at far greater distances than an aircraft to a ground station. A simple call to any listening station or aircraft on 121.5 requesting assistance will almost always get a response.
If VHF communication is not possible, you could try getting a patch through to ATC on HF radio. My favorite back in the 1990s was Stockholm radio. These days, however, SATCOM is available. I don't know anything about it, but I imagine it provides the ability of an aircraft anyplace in the world to be patched through to an appropriate ATC facility.
So, if you're talking to ATC (by whatever means) but you're out of radar coverage, what then? Every captain would remember generally where he was prior to the failure. It could be as simple as someplace in the North Atlantic and eastbound. He could simply tell ATC that and that he will fly a westerly heading. ATC would clear the altitude he was at and look for him along the westerly limits of their radar coverage for when he inevitably would show up. Actually if it was the North Atlantic and it was nighttime, he could just follow any airplane in front of him. They're easy to see stacked out into the distance.
The general idea would be to get to where you could be seen by ATC, and go from there. Anymore, there aren't a lot of areas along regularly traveled airline routes over land in North America and Europe where above 30,000 ft an airline flight won't be seen. That was true back in the 1990s, more true now I am sure, and increasingly true through the world, especially if military radar can be utilized..
The more interesting problem becomes that there is a loss of communication as well as a loss of navigational ability. The possibilities here are limitless. A simplification would be to say that what you would do would be to take up a heading that you believe will get you to where you could visibly see a land mass with airports, and land at the first suitable airport you see. Most airline flights are along routes that the crew is very familiar with; they know what they've been flying over many times, and that would be a help.

SATCOM is available almost everywhere -- the geostationary Inmarsat birds can't be reached from the geographic poles.  (You should still be able to raise someone on HF from there, though.)

A couple other things...first, you probably ought to mention that "nearest suitable" would involve either VMC or the availability of a GCA (ASR or PAR) approach (since without working nav systems, you wouldn't be able to shoot a normal instrument approach).  Second, there is an off chance that someone'd be sent to intercept you under those conditions (which is a good thing, because you can simply then form up on the interceptor's wing and let them lead you to a safe place to land).

There are several different navigation systems in commercial airliners that can operate independently of each other such as:

GPS
INS (Inertia Navigational System)
VOR

and more besides.
The chances of all these systems failing simultaneously are so slim it's not normally considered. However the assuming other instruments are functional there are still methods of navigation available.
To start with, you would probably want to talk to air traffic control and advise them of your situation. If there is radar coverage for your location (probable if over land) they can track your position and keep you updated as required.
Secondly, you would know your position at some point prior to the system failure, your airspeed and direction. So you could use this information to work out your approximate position.
Finally if you're over land you could use good old fashioned navigation using landmarks to work out where you are and which way to go.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Gas Venting Stations: What's The Big Deal?
On sectional charts you can see that, through various restrictions, regulators encourage pilots to avoid flying too close to Gas Venting Stations. These seem to be places where natural gas may occasionally be vented into the air (for example from pipelines). I have been unable to determine if the gas is preignited.
Is the principal risk from gas discharge that's being mitigated here:

aerodynamic
explosive/thermal
respiratory
a combination of the above?

Have incidents occurred relating to these risks?
You can find some talk of the risks online but it seems mainly to be in the form of "My mate told me...".

Can you provide an example?

49 examples on this page: look for GVS. https://www.aurora.nats.co.uk/htmlAIP/Publications/2020-01-02-AIRAC/html/eAIP/EG-ENR-5.3-en-GB.html

Would help to add the UK tag, we don't have those types of advisories in the US.

I'm not sure why whether it happens in the USA or not is particularly relevant, but if it's UK-specific that's a fair point.

There's a comment by the Kidderminster GVS that says "Strong thermals, smoke and gases" which may be a clue as to the nature of the hazards.

It's not UK specific. I've seen similar warnings on Dutch approach plates, and I am sure other countries have them too.

Gas venting stations sometimes release natural gas at extremely high pressures, so there is a risk of strong updrafts or turbulence. In any case if you fly directly over you may be going through a cloud of combustable gas which could possible ignite. It's a low risk, but given it's rarely a problem going around them it would be my advice to do so.
I've flown over one without realizing and it put me in a 1200ft climb, I had to cut throttle in a hurry and dive to stay out of the London TMA. After that I paid attention to them.

That sounds surprisingly violent. May I ask which aircraft you were flying and at which altitude?

It was in a Cessna 152, IIRC 1800ft in VMC. It wasn't particularly violent,  but I felt it. By the time I figured it out i was  about 200 feet below a class A.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What technology is being moved forward to keep pilots from crashing planes?
So it seems that we have three basic issues (and possibly more but not as prevalent):

Pilots can crash planes.  Someone can just be suicidal, be part of an extremist/terrorist group, can have a vendetta against someone or just thinks might as well take more down with me.
Passengers that get into the cockpits.  I know that the cockpits are locked and pilots are armed but on a lot of flights there is opportunity.
Using technology to override aircraft autopilot.

So what is the long-term plan to keep these things from happening from a technology point of view?  Is there anything in the makes that will help solve these issues?  I foresee some sort of technology that requires some sort of dual sequenced unlocking to allow a pilot to make a certain % deviation from flight plan.
Note:  There have been a lot of answers here that basically say "nothing is being done" and offer up reasons why.  To use the recent incident of a pilot dropping a plane into a mountain as an example, my main question surrounding that is, is there a technology that would see that the pilot was well off course?  Is there technology that would see that the plane had a high risk of crashing?  Is there technology that would prewarn flight control (even if it were a few minutes)?  Is there technology that flight control could override plane and right the ship that is going into the mountain?

And what if a pilot has a medical emergency and they need to divert? There is no technical solution to having a pilot who has the ability to deal with emergencies while preventing them from crashing the plane.

@cpast - I am not say that the dual unlock is just two people, it would just take two people to unlock.  (copilot and airline or two airline ground employees).  Whatever.  I don't have the answers but it seems like this should be something that is going to change.

Pilots crashing planes intentionally is so ridiculously rare there is no need to create measures against that. And it would be a completely reactive, not preventive, measure: just like the TSA operates (oh, that one time they tried to blow up the plane using liquids, let's ban liquids, although there are virtually endless other possibilities on how it could be blown up, but we will start caring about them the first time they try)

Wouldn't that introduce a possibility that Air Marshal will crash the plane?. If we can trust the pilot, why would we trust someone else? This comment only to illustrate the question is a dead end.

I do not understand why I would trust the marshal any more than I would trust the pilot but supposing I did, why a stun gun?  Surely lethal force in such situations would be called for?

@AndreasBonini - well it only takes two of these incidents to put an airline out of business and there are a lot of copycat reactions in the world... so I would think this is something major airlines would be looking into.

@Simon lethal force is never called for if it can be avoided. If an Air Marshal can prevent a pilot from posing a danger to the aircraft and its passengers without harming the pilot, that must always be preferred.

OK, so you want to put a political officer in charge of the pilots. What happens when the political officer forces the pilots to make a bad decision? Everyone dies.

Note that media sensationalization does not imply a sudden increase in a problem or sudden increased need to solve it (nor does it create an increase in benefit in a cost-benefit analysis of a solution). Just because you are suddenly aware of it (and care about it, for at least a few more days/weeks until the next big story comes up) does not mean it's an increased danger. It just means your favorite news anchors have something to talk about to keep you watching and keep their twitter feeds active for a few days.

I think only in the US pilots are armed, at least not in Europe.

@raptortech97 I burned the 'crazy' part. It was offensive and superfluous.

Every person needs to be taught basic risk management in high school or something.  There are TWO components to risk management:  frequency and severity.  You multiply them together to determine actual risk.  An airplane crash, while extremely severe is very rare. Conversely, auto accidents are less severe (only a few people die at a time), but they happen all the time.  As a result, we know that riding in a car as dramatically more risky than flying in a plane.

Two people can override the pilot's lockout?  Great, now an attacker can just force two people to operate the override and can storm the cockpit.  @Simon seems like a stun gun would be safer, stray bullet could easily puncture the plane's skin and cause serious issues.

To answer the new questions in the edit: 1) Yes, it's called "radar." 2) Yes, it's called "pilots." 3) Yes, it's called "pilots if they can spare the time," and sometimes "air traffic controllers" (if the flight track reflects the issues). 4) No, and it's actually a pretty bad idea to do this -- a pilot has to have the ability to deal with emergencies, and they need not talk to ATC when they're doing so (the order is aviate, navigate, then communicate). You can't tell intent through sensors, and a conscious non-malicious pilot is better at dealing with emergencies than ground control.

@Michael A bullet through the fuselage would hardly be noticed.  A bit of a whistling noise maybe but aircraft can cope with holes very much larger than a bullet.  It is almost impossible to significantly damage an aircraft with a bullet, especially from a handgun.  You might want to review this question

Nothing.
At present, there are no serious technological plans to prevent pilots from intentionally crashing a plane. There are discussions about making sure that there are always two people in the cockpit, and at some point policies might be changed to make it easier for pilots to enter the cockpit (even if no-one in the cockpit wants them to), but ultimately we trust the pilot to do whatever necessary to ensure the safety of the aircraft (within certain bounds; see the flight envelope protection system in Airbus aircraft).
By limiting pilots' ability to control their aircraft, we seriously limit their ability to legitimately deal with emergencies. See, for example, cpast's example of one pilot having a medical emergency and the other pilot not being able to divert because they can't get approval. Even if both pilots are conscious and able, such a system would slow them down when responding to emergencies.

Well, I'd argue that there are lots of technological means to attempt to prevent pilots from crashing planes. Pitch stability, stick shakers, stall horns, lots of fancy warning lights and messages, the entire air traffic control system, etc. all come to mind. :) There just aren't many measures to prevent pilots from intentionally crashing planes, aside from the Air Force potentially having some say about where they crash them.

@reirab good point! In a way, even the wheels are an anti-crashing system!

How do you know there are no technological plans?  I know for instance that there is a lot of talk about automating space flights, so wouldn't they pass on the technology?  I really don't like that the most upvoted answer just says, nothing, with no evidence whatsoever.

@blankip I can't tell you that nothing is happening. All I can tell you is that any such technology would be ridiculous and dangerous, and would be the result of a radically different attitude towards pilots than the current norm. I'm not sure what you're talking about when you say "automating space flights." A lot of flying is already automated, and it's likely that trend will continue. It may be that at some time in the future, pilots won't hand-fly at all. Whatever the case, though, pilots will always ultimately remain in control of their aircraft. At least for the foreseeable future.

Until there is a technology that logically passes control to the pilot in a safe way and limits the pilot based on rules....  I am not saying pilots shouldn't have control at all.  I am saying that there is no reason a pilot should be able to divert 100s of miles off course with no confirmation.

@blankip I don't see why they shouldn't be able to. What if the hydraulics are damaged, and they can't bank? What if they need to divert?

Then there are rules set in place for the amount of diversion before flight control needs to give the pilot complete override (and maybe confirming with co-pilot).

@blankip what do you mean by "complete override"? And remember the plane should still be flyable if one pilot falls sick or is otherwise able to confirm or verify anything.

Yes then you could move to flight attendant or set whatever protocol.  Laying out a series of complex protocols would certainly catch most issues.  Also I would think a plane should know that if it is nose diving towards a mountain at 3K feet that it needs to do something else to stay alive.

@blankip It's not that simple. Please remember that adding non-overridable security protocols (secure cockpit door with in-cockpit lockout button) is what got us into this particular situation in the first place. Adding "complex protocols" may protect against a particular kind of threat, but it will open up the possibility of an entirely different risk altogether ("an entirely different risk"). Any system that can't somehow be disabled on an aircraft is dangerous; you can't pull its circuit breaker if there's a fire or malfunction. And what if the override doesn't work and you need to divert?

Already this tragedy was caused by another security change. Before 9/11, the main pilot could have enter the cockpit, or at least could have crashed the door, because it wasn't reinforced against a possible terrorist attack.

"Complex protocols" that have to be satisfied before a pilot can deviate from the original flight plan will make the vast majority of cases more dangerous, not less. The proportion of flights that are deliberately crashed is insignificant compared to the proportion that experience malfunction, unexpected weather changes, or have to respond to some external factor like bird strike, unplanned redirects etc. In every one of those cases, complex protocols slow the pilots down in a situation where a fast response saves lives.

@blankip it's also worth noting that the aircraft would have "known" that it was heading for a collision. Many flight instrument systems will have a warning in place, using GPS terrain maps and/or terrain avoidance radar, which will display a warning and play an audible alert stating "terrain", "altitude", or even just "pull up". If you're on autopilot, they will respond to this for you, but if you're flying manually they won't cut in and take over because in the vast majority of cases there's a good reason you're flying manually, so it's assumed to be not safe for the autopilot to take over.

You can't. There exists no technical solution to this problem, aside from removing pilots entirely from flight decks. Pilotless aircraft would prevent this, but that's not "stopping the pilot from crashing the plane," it's removing the pilot entirely (and is then more vulnerable to unforseen situations, unless there's ground control, where you have the same problem where a person can crash the plane).
You have to pick what will have ultimate control of the aircraft, knowing that if it messes up, you will crash. Something must have ultimate, non-overridable control. It can be an autopilot or a human or a bunch of humans. Autopilots are much less flexible than humans, making them an imperfect choice for the controller-of-last-resort (even in Airbuses, where the flight envelope protection normally restricts the pilot, it hands control over to them if there's an issue it can't solve, and the pilot can turn it off by pulling enough circuit breakers). If a human has that role, they can crash the plane. If you want two people to have to share that role, you have to deal with what happens when you only have one available and need to act now. If one person has the role, they can crash the plane.
Not all problems have technical solutions. Aircraft are designed with the asusmption that a human pilot is in charge, and that they know what they're doing and can deal with abnormal situations. The aircraft can't always tell if there's an abnormal situation; no one on the ground really can either. Trying to have pilots who can't crash the plane is like trying to keep a doctor from murdering a patient (you can't do so through technical means, because no formal rule can perfectly sort out when something is a legitimate response to a problem and when it's not). The only solution is to make sure that pilots aren't the type to intentionally crash planes. Which, I mean, they do a pretty good job of -- intentionally crashing planes is incredibly uncommon among commercial pilots.

I do want to add there's another thing you can do - have another pilot in the cockpit, and also have flight attendants. See, for example, JetBlue 191.

Wouldn't having the ultimate control at flight control with a group of people create a more ideal override situation?  I would feel better if my pilot could only deviate if a group of people in flight control didn't allow him.  Also I would think there would be technology that could help revert a plane from crashing into something as big as a mountain.  Your answer seems more about what you can't think of than what might be a solution in the future.

@blankip "Intentionally crashing planes is incredibly uncommon among commercial pilots." Media sensationalization does not imply a sudden increase in a problem or sudden increased need to solve it. Just because you are suddenly aware of it (and care about it, for at least a few more weeks until the next big story comes up), does not mean it's suddenly an increased danger. It also does not mean it's something that hasn't been thought about and considered before.

@blankip "Wouldn't having the ultimate control at flight control with a group of people create a more ideal override situation?" Have you ever sat in a meeting at a large corporation with a lot of management people and tried to get a decision made? No one ever wants to take responsibility for a simple decision, and you want a committee to decide the fate of 200 lives when seconds count?

Heh, I love the phrase "by pulling enough circuit breakers." Just pull every breaker, one by one, until you get control back.

Systems exist to avoid CFIT (controlled flight into terrain). This has been the leading cause of fatalities in the last 50 years: Intact planes being flown into the ground or mountain sides by disoriented pilots and wrongly entered data. Precise navigation and detailed terrain maps make it possible that the F-16 will not allow its pilot to fly into terrain. Currently, the system on airliners will only warn that ground contact is imminent, but not prevent it. Enabling prevention is long overdue and entirely possible in airliners, and I expect both manufacturers, Airbus and Boeing, to expedite their plans to get such a system certified for adoption into regular airliners. Maybe it will even be made mandatory.
This will at least allow more time in cases like that of 4U 9525, but it could not entirely prevent them. Just take the Mount Salak crash: The pilots thought they had a database problem and disabled the warnings. The system must enable ground contact when near to airports, and cannot avoid it when the navigation subsystem fails (no GPS contact plus wrong starting coordinates). In the end this kind of thinking reminds me of all those TSA tactics which always focus on preventing the last incident instead of avoiding the next one.
In the end, if people are determined enough, they will find a way around all preventive measures. Human ingenuity will always trump bureaucratic procedures.

Also, pilots must always have the authority to make an emergency landing at a non-airport.

@raptortech97: Good point, but this should only be needed in case of engine or complete nav system failure. The system can only prevent ground contact if enough thrust is available, and would minimize sink rate otherwise. Ground contact will then be inevitable, and the pilot should be able to select the precise spot.

Which, of course, probably wouldn't stop a pilot from turning off the nav equipment or shutting off the engines to disable the system. The point of all these things is more to prevent accidents than to prevent intentional crashing.

"In the end, if people are determined enough, they will find a way around all preventive measures. Human ingenuity will always trump bureaucratic procedures." That right there is the winning statement. If a human can devise a solution, another human will come up with a way to defeat it.

Sometimes in software engineering you'll hear people say something like "We built this system to be idiot proof, and then they built better idiots."

First two items in your list are part of class of crash issues so called "controlled flight into terrain" which is part of "human factor" issues.
The only way to avoid such crashes is to create airplane without pilots at all, that is what Google try to do with automobiles. And I believe there are no technical problems to do it right now. Modern aircrafts are able to takeoff, flight and land in fully automatic mode right now and do it in much safer manner than most pilots.
However there is social problem, we do not trust computers that much and most people prefere to fly with human pilot in command rather than a autopilot.

"Modern aircrafts are able to takeoff, flight and land in fully automatic mode": Yes... most of the time. What about A/P disconnecting when fed with inaccurate air data (AF-447)? In addition this is not true. There is no airliner which can operate landing gears automatically, at least not a civil one.

No, it's not a "social problem".  It's a well reasoned approach to risk management, and a mature awareness of what happens when technology fails.  Technology is not magic from the gods.  It is built by people who try to anticipate all the things that can go wrong with a system.  However, if they miss something, and the system ends up in an undefined state, your best chance at not dying is with a human being improvising using their experience and knowledge as a guide.

Other than background checks and rules of practice about who is allowed to fly what when and who can/must be in the cockpit, no technical measures are currently being moved forward for that.
I don't think I'd want to have such technical measures in place, either.  Sometimes, a pilot needs to crash-land an airplane in the best interests of public safety.  The "Miracle on the Hudson" is probably the best example of this - the pilot could have tried to get to a landing strip but would likely have tragically crashed in a populated area; instead the pilot chose to crash-land the plane by flying it into terrain in a particular way that fortunately led to no loss of life by anybody on the plane or ground.
Flight 93 (to the extent that the general population now accurately understands the story of what happened on the 4th 9/11 jet) is another good example of when flying the plane directly into the ground, even if that kills everybody on board, was better than an alternative which would have probably killed everybody on board and a whole lot more.
Especially if there is a loss of cabin pressure or other emergency where a pilot needs to descend rapidly and change the route, having the authority to do that is important.

Requiring two pilots in the cockpit at all times.
I am not sure that counts as a technological solution however.
You could imagine they could put emergency intercoms to air traffic control outside the cockpit and allow air traffic control to take control of an aircraft in an emergency.  Or at least to remotely open the cockpit door.
Or... you could let the passengers vote electronically for whether the cockpit door should be opened or not...


text:
What is the effect of flying height on helicopter fuel consumption?
Does the height at which a helicopter is flying affect its fuel consumption? Which factors are influencing its fuel consumption?

Yes, altitude affects fuel burnt. See other elements page 24 of the HAI Guide For The Presentation Of Helicopter Operating Cost Estimates

The lift equation shows that as air density decreases, as it does with altitude, in order to maintain lift, the speed of the blades must increase or, the coefficient of lift (CL) must increase.  Since the characteristics of the blades do not change, in order to increase the CL, the angle of attack must be increased.
Increasing either of these increases drag so, in order to prevent the blades from slowing down, the engine must produce more power to counteract the drag.
Most helicopters use constant speed rotors so in practice, the angle of attack (the pitch of the blades) is the only variable.  In order to maintain speed of the blades, the pilot, or a throttle governor, must increase power from the engine.
Therefore, a helicopter uses more fuel as it's altitude increases.

what helis don't have constant speed rotors?

@rbp The A160 for example, but I'm jut covering my behind lest some smart Alec says "what about the Acme Wing Flinger 40?"

H260 not collective-controlled  you can definitely fly a heli with low rotor rom but it's not a normal operations

@rbp Of course, I was just trying to keep it simple for the lay person leading to collective pitch and engine power being the variables.  My worst experience ever, lesson learned and never repeated, was pulling collective before I'd joined the needles at the end of a power recovery auto. Good job my instructor was on the ball.  I personally learned how much below NrMin you can go in an R22 :(


text:
Does lift equal weight in a climb?
This subject keeps coming up in the discussions and questions such as this one, which asks if lift equals weight in level flight. Good answers there, pointing out that upwards force has many sources. But also some that need clarification.

It is also mentioned at several places on this Aviation SE site, in question & comments, that lift always equals weight if the aeroplane is not accelerating upwards or downwards, since only an acceleration requires extra force according to Newton.
With zero wind, lift is always defined as the force perpendicular to the flight path, but gravity does not tilt with the aircraft axes. My question therefore is also about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft.

If larger: please quantify.
If equal: please explain why.

I would contest that list is not always defined as a force perpendicular to the flight path.

What is the definition of lift?

Related: https://aviation.stackexchange.com/questions/12714/why-do-airplanes-lift-up-their-nose-to-climb

@JonathanWalters: Right, wind will add an airspeed component that might not be aligned with the flight path.

Included in the question.

The picture shows a NASA DC-9 in a parabolic flight trajectory. This is not an example of steady climb. Working from the chart on page 334 of http://pcmuseum.tripod.com/dc3/dc3performanceinformation.pdf I get a flight path angle of about 8 degrees for a steady climb from sea level, very different from the FPA in the photo.

Yes indeed. Great picture though.

"there is now a vertical aerodynamic drag component with needs to be overcome by aerodynamic lift" But this is wrong. Climbing requires less lift. The additional upwards force to overcome drag comes from thrust. In fact, any attempt to climb without reducing lift can not result in a statically stable configuration.

No the aircraft is still level, thrust is still horizontal and does not contribute to vertical force. The aircraft climbs by flying faster, not by pointing the nose up.

Where the nose is pointing is irrelevant. What's relevant is the thrust vector, which in many aircraft (but not thrust vectored aircraft like the Harrier or Osprey, or hellicopters, or most spacecraft) just happens to be fixed with respect to the nose. By the usual definitions, you can not overcome drag with lift because drag is by definition orthogonal to lift. Thus, drag can only be overcome with thrust. Of course you are free to redefine terms however you like, but the definitions you are using do not seem to be compatible with the conventional definitions.

I'm sorry Phil, you've lost me. How am I redefining lift?

You are supposing that additional drag can be overcome by additional lift. But this is not possible under the conventional definition where lift is always orthogonal to lift. If lift is "up" by definition, then drag is "to the rear". You can't offset a force pushing to the rear by pushing up.

Additional horizontal drag is overcome by additional horizontal thrust. The aircraft is flying faster, without changing attitude. The air is flowing faster over the wing, in doing so increasing lift.

Let us continue this discussion in chat.

In your example the climb angle can only be as large as the difference in angle of attack between slow, horizontal flight and fast, climbing flight. This is not much and completely unrealistic. Of course you need to change attitude to allow the aircraft to climb. And still, even then the lift is pointing slightly backwards and is lower in magnitude than the weight. BTW, the accepted answer does not answer the question but states a physical tautology.

This question needs to be improved.  The title does not match the body of the question.  The sixth sentence is "My question is purely about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft."   That is not a question about lift, that is a question about net vertical force.  The title is misleading.

@quietflyer Indeed, have updated.

Given you have already self answered this question I would suggest removing the answer/update section from the question as per this meta post

@Kevin would you mind explaining why you removed the flight-mechanics tag that I added?  It seems it would fit here as well as anywhere else.

It is unfortunate that this question got edited to assert that the propeller-train analogy implies that lift is greater than weight in a climb.  This question would be improved by deleting all references to the propeller train.  This answer explains why the propeller-train is not a good analog for normal free flight -- https://aviation.stackexchange.com/questions/15243/is-excess-lift-or-excess-power-needed-for-a-climb/57034#57034

Now 1 more-- now there are currently 8 deleted posts connected to this thread-- I wonder if that is an ASE record?

F=ma, dude.... F=ma. With that said, every other explanation is just wordsmithing or redefining terms. Total force must be zero if the acceleration is zero. So first decide how you want to define your terms, then answer the question. But if your answer is no, then you better have defined some other force term to make up the difference, because if the acceleration is zero, then the sum of forces must be zero.

@CharlesBretana F=ma dude, totally agree. If an F16 flies straight up at a constant velocity, is the thrust larger than weight?

@Phil Frost, what frame of reference are you using when you say lift is by definition always “up”?

This subject is rapidly becoming the next “if an airplane was on a treadmill...”

@MichaelHall I wrote "If lift is "up" by definition". Whatever frame of reference you choose, lift and drag are typically orthogonal. So if you choose a frame of reference where lift is up, then drag is to the rear.

It depends on exactly how you define "lift" and "weight". You might say intuitively that lift is all the forces acting on the aircraft in the upward direction, like this:

In this case, lift must equal weight, otherwise the aircraft would be accelerating. That is, it's rate of climb would be changing.
But it's more usual to define lift this way:

Here, lift and weight are equal in magnitude, but in different directions. Of course lift doesn't need to be equal in magnitude: it can be adjusted by the angle of attack. But let's suppose lift is equal to weight and see what happens.
Let's do all our calculations with Earth as the frame of reference1. It's useful to decompose lift into a sum of vertical and horizontal components so we can analyze the horizontal forces and the vertical forces separately:

Comparing the vertical component of lift with weight, we can see they are not equal:

Considering only the vertical forces drawn here, there is a net downward force on the aircraft. So why then is the rate of climb not decreasing?
A similar transformation happens to thrust. In a climb, thrust provides an additional upwards component. And of course we must also consider drag. Point being in a steady climb, lift (by the conventional definition) is not equal to weight, but the sum of all the vertical components of lift, thrust, and drag do equal weight.
Let's add an arbitrary amount of drag, and enough thrust to balance the vertical forces.

Now the vertical forces are balanced, but the horizontal forces must also be balanced if we want stable flight. Adding all the horizontal forces in my drawing, there's a net force to the left. So this aircraft may be maintaining a steady rate of climb at this instant, but it's losing speed and probably headed for a stall.

Remember, we initially set lift equal in magnitude to weight, and this is what happens. Without changing the direction or magnitude of lift, there's no solution that results in stable flight.
Therefore, a climbing aircraft requires less lift. To maintain this direction and velocity, this pilot must reduce lift by reducing the angle of attack, and increase thrust such that the vectors add to zero and there's no net force on the aircraft. Reducing lift will also reduce drag.

1 Any other frame of reference could work. For example we could use the aircraft as the frame of reference, which would mean lift is always up, but weight would change direction.

So we're transforming from an aircraft frame of reference to an earth frame of reference, correct?

You could do both. When you use the aircraft frame of reference in a climb, aircraft down is not aligned with gravity. The gravity vector in that case would then point slightly towards the tail (as much as the lift vector is from the earth as a point of reference). Using earth frame of reference might be more common though since gravity vector then is mostly constant.

I don't think there's any particular reason why the lift in your second image would have to have the same magnitude as the weight. After all, it can be modulated by varying the pitch (and so the angle of attack), just as in horizontal flight.

@HenningMakholm Right. I just arbitrarily picked something to illustrate.

@Koyovis Yes. I added a footnote to clarify.

Perhaps you should edit your answer to reflect Makholm's comment - right now, the answer can be very confusing, leading readers to believe lift should always equal weight.

@Sanchises Better?

On the other hand, if climbing like this without wings much more effort must be expended than if the wings were there. As the thrust goes up so does the speed and therefore the lift until forced back almost to the steady state form. What happened to this setup is once traveling fast enough the lift exceeds the weight.

@Joshua Not sure I follow...was that a question?

I wonder how many aircraft (other than military aircraft and some specialized aircraft) can maintain a steady climb at the flight path angle shown in the figures. I'd say the first thing to do (in an "ordinary" airplane) would be to reduce the flight path angle (and thereby also reduce the rate of climb at a given airspeed).

I'm interested by your observation that less lift is needed during climb due to the backward pointing lift vector. This makes sense but as it's due to the AoA, why lift the nose to climb? Will increased thrust not increase lift more effectively at low AoA by your observation, and so lead to more effective climb? To be clear, I am not doubting you, I am trying to understand.

@DanSheppard Keep in mind AoA is the angle between the wing and the relative wind, which isn't necessarily the same as the pitch. If the climb rate increases but the pitch doesn't increase, AoA decreases because the wind appears to be coming from above the wing. So to attain a reasonable climb rate, the nose must pitch up. Consider also: if you think of these vectors with the airframe (rather than the Earth) as your frame of reference, then in a climb gravity is pulling a little less down (so you need less lift) and a little more back (so you need more thrust). That's perhaps more intuitive.

Why does the first drawing only depict gravity as a vertical downforce? In a climb, there is a vertical component of airspeed, creating a vertical component of aerodynamic resistance. Purely regarding the question from an earth axes reference frame. Intuitively clear when a fighter jet or a helicopter flies vertically upward.

@Koyovis I don't understand your last two sentences. I didn't rigorously define my coordinate system besides "Earth as the frame of reference", but wouldn't it make sense for this coordinate system to define "down" precisely as "the direction gravity pulls"?

Yes it does, we're in agreement there. It's just that the vertical downwards air resistance force is missing in the picture.

@Koyovis It's been so long since I wrote this, but I guess since the initial assumption of "lift equals weight" doesn't include drag or thrust, I didn't start with that either. But they do get added later in the explanation.

In an aircraft that is climbing at a constant vertical velocity, the total of the upward-directed vertical forces is the same as the total of the downward-directed vertical forces.
Were it not so, the vertical velocity would not be constant, since any non-zero balance of the vertical forces would result in an acceleration...

By total of downward forces, do you mean the weight?

@Koyovis  Not only the weight. The downward component of drag is a force to be reckoned with too...

Which means that the total upward force is greater than the weight if the airplane is rising, because, as a bare minimum, there is always a downward component to drag in addition to the weight in that case.

@MadPhysicist Exactly right. Lift vector tilts back, therefore drag vector tilts back, creating a downwards pointing component of aerodynamic drag.

All "Forces" are just engineering definitions to allow calculations to be done in a simpler manner. The only real forces are the thrust of the engine, and the pressure force of the air hitting the surface of the airframe.  Even Gravity is a "fictitious" force, in that it is only necessary to balance the equations because we are measuring everything in an accelerated frame of reference (accelerating upwards at 1 "G"

@CharlesBretana All forces? Just engineering definitions to simplify calculations? I'd like to see you do any calculations at all with no forces, even in an inertial reference frame.

Read what I said.  "All "Forces" are just engineering definitions that allow calculations to be done in a simpler manner"... What are you quibbling about? If you shoot a bullet up in the air, does it have a real, actual, physical horizontal velocity, and a real actual vertical velocity ? Of course not, those are both fictitious, (but extremely useful) simplifications to allow us to do calculations. But they are still just engineering definitions, fictions.

@CharlesBretana Well, Zeno of Elea had an argument which, if it were sound, would prove your assertion about bullets (though he considered arrows instead), but recently (at least since Aristotle), there have been doubts about its validity.

@sdenham, my apologies for seeming to be "preaching"? a bit. This issue is one I find to be the cause of so much misunderstanding that when I see it I find myself irresistibly drawn to attempt to clarify it. Almost every mathematical equation, variable, expression or concept is an abstraction, whose validity, as you put it, is true only within carefully defined constraints and conditions, the first of which, is generally defined by specifying the particular frame of reference (by position, velocity and acceleration), in which it is being considered.

@CharlesBretana "Almost every mathematical equation, variable, expression or concept is an abstraction." Almost? how do you determine which ones are not - and which ones are they, anyway? Is a frame of reference not an abstraction? The question of what's really real is a vexing one, and discussions of it always seem to end up with a subjective choice of axioms.... BTW, validity, as I used it, is a property of arguments, not equations, variables, expressions or  concepts.

Hah!  At the moment, I can't think of one that is not an abstraction. Indeed, at every point in the human history of science, when we have had a paradigm shift, a conceptual revolution or an expansion of our understanding, it has included a realization that the objects or concepts that we previously thought were the real fundamental building blocks of reality are actually just constructed mathematical aggregate abstractions of slightly more fundamental objects or concepts at a deeper level.

"How does one tell?"  Well, just off the top of my head, one clue might be when generally accepted peer-reviewed scientific theories describe an object as being composed of smaller objects (like protons & neutrons) or when a concept or engineering variable or value is described as the result of some calculation or combination of other values, variables or objects (like Pressure, or accelerations (the rate of change of an objects measured velocity), or Velocities, (the rate of change of an objects measured position) or temperature (the average velocity of atoms and molecules in matter)

Short answer: No.
Long answer: When the flight path is not horizontal, lift will not be vertical but perpendicular to the direction of motion (in still air). Thrust will also have a vertical component and is different in magnitude from drag, because excess thrust is needed to increase the potential energy of the plane.  Note that the vertical component of lift is proportional to the cosine of the flight path angle while the vertical component of thrust is proportional to the sine of the flight path angle, so the thrust part grows more quickly at small flight path angles. Therefore, when climbing, thrust will add some vertical component, so less lift is needed.
Again, in a descent less lift is needed. Now thrust is smaller than drag, and drag, pointing slightly upwards, contributes a vertical component, counteracting weight. So in both cases lift is smaller than weight.
So far, this has been unaccelerated flight. But normally a climb has acceleration components:

to adjust speed to the change in density (accelerated in order to stay at the same indicated airspeed) or Mach number (decelerated in order to stay at the same Mach number), and
because the aircraft loses vertical speed as thrust is diminished by the change in density and, in case of propeller aircraft and turbofans, by the increase in true airspeed.

This second, admittedly tiny, effect will add a vertical inertial force which adds to the remaining vertical forces, namely lift and thrust. When this inertial force is considered, the remaining vertical forces are a tiny bit lower than weight.

Likewise, when descending, drag will add some vertical component to oppose gravity. The lift component for either climb or descent will be equal to weight x cosine(angle from horizontal). However in the second part of the original question, if the aircraft is not accelerating, then the net force on the aircraft is zero. Gravity would be opposed by some combination of lift and/or thrust, and/or drag.

@HenningMakholm: I linked to an answer which has a diagram quite like you wish. I was told not to repeat myself and to better link to existing answers.

Is vertical component of thrust the sine of the flight path angle or of aircraft attitude relative to earth?

@Koyovis: Only if thrust is aligned with the aircraft's longitudinal axis. Roughly it is, but there can be a few degree of difference.

If we define lift as the component of the total aerodynamic forces on the aircraft that is perpendicular to its direction of motion, then lift will be slightly smaller in a stable climb.
It is probably easiest to analyze the situation in a coordinate system that is tilted such that one of the axes is parallel to the direction of motion. Then all of the forces -- lift, drag, thrust -- work just like in an ordinary coordinate system in horizontal flight. The only difference is that the weight force now has a different direction -- but still the same magnitude.
This means that the component of weight that is perpendicular to the motion is now slightly smaller, and lift must be correspondingly smaller too. The plane's angle of attack will be slightly smaller than in level flight at the same (calibrated) airspeed.
On the other hand, the weight vector now gains a significant component parallel to the direction of motion, and this has to be counteracted by more thrust, lest the aircraft would slow down.  (This will much dominate over the small decrease in induced drag that results from the slightly smaller lift).

Yes seen from the aircraft reference frame, weight now decomposes into lift and drag.

TL;DNR
Does lift equal weight in a steady state climb? Vertical force is higher in a steady state climb, but lift may be tilted depending on how the climb is executed, is the answer. Tilt the aircraft axis relative to earth axes, and by definition part of the gravity vector is now in the aeroplane thrust/drag axis. That is very clear, and the case that everyone refers to in their answers in great detail. However, a steady state climb can also be executed with the nose pointing straight forward, and then lift is larger than weight. And helicopters are aircraft as well...
Full answer
It depends on the relative axis orientation.

Gravity is always aligned with earth axes.
For fixed wing aircraft, lift and drag are aligned with the airflow axes (aligned with the airflow in a steady state starting position). Note that thrust is only aligned with drag at AoA zero.

The thing is that for fixed wing aircraft, a steady state climb is mostly and automatically associated with increasing AoA, which tilts the aircraft axes upwards, resulting in an upward tilt of the airflow axes. But fixed wing aircraft can also climb by increasing speed, which results in a steady state climb with a reduced AoA.
Below is an analysis of the two cases for fixed wing climb, and for helicopters where the airflow axes are rotating with the blades - providing lift, not thrust.
TL;DNR

Fixed wing climb by increasing AoA: modulus of the lift vector < gravity vector
Fixed wing climb by increasing velocity: mod lift > mod g
Helicopter in steady climb: mod lift >> mod g

Lift L at angle $\alpha$
Drag D at angle $\alpha$
Thrust T at angle $\phi$
Weight W at the vertical

Force equilibrium in unaccelerated flight:
$$ T\cdot cos(\phi) = L\cdot sin(\alpha) + D\cdot cos(\alpha) \tag{H} $$
$$ L\cdot cos(\alpha) + T\cdot sin(\phi) = D \cdot sin(\alpha) + W \tag{V} $$
Equation (V) states that the total upwards vertical force is equal to weight plus a component of aerodynamic drag - of the whole aircraft, wing + fuselage + tail etc. So the total upwards force will always be greater than weight, unless $\alpha$ = 0
Let's have a look at a couple of cases.
1. Climb due to increase in speed, fixed wing
A case pointed out some time ago by Chris, who defined totally uncoupled thrust and lift forces by putting a wing on a pole mounted on a train car. If thrust increases, speed will increase and the wing will climb upwards with a constant velocity $V_z$. This will change angle of attack, and will tilt the lift vector backwards. The wing climbs with constant velocity once the total upwards vertical force is identical to weight, plus the vertical component of drag which points downwards.

Note that thrust is nowhere to be seen in this picture, only aerodynamic forces. Thrust is set at angle $\phi$ = 0 and will be equal to L * sin($\alpha$) + D * cos ($\alpha$). Lift L is tilted backwards by angle $\alpha$, and is greater than the vertical upward force by factor $1/cos (\alpha) $.
So in this case (climb by increase in speed):

Total upward force is greater than weight by an amount of D * sin $\alpha$.
Lift is the only contributor to upward force, is tilted back, and is greater than total upward vertical force.

2. Climb due to aircraft pitch up, fixed wing
Now let's have a closer look at the case of a fixed wing aircraft, climbing because of an increase in pitch angle. All the above forces and both equation (H) and (V) are to be considered. Angle of attack $\alpha$ is defined by pitch angle $\phi$, airspeed V, and climb speed $\dot{z}$.
So in this case:

Total upward force is again greater than weight by an amount of D * sin($\alpha$)
Both thrust T and lift L are contributors to total upward force. How much each contributes depends on pitch angle $\phi$ and climb speed $\dot{z}$. More axes tilt means: smaller proportion of Lift, larger proportion of Thrust.

3. Helicopter in vertical climb
Now for the helicopter in climb. At first glance, this is a case of only thrust being responsible for the climb action, because the rotor disk delivers vertical thrust downwards. But here is the thing: that is from a fuselage perspective, but now lift is defined relative to the rotating blade airspeed.
Our reference frame is once again earth axes. The vertically climbing helicopter has the same downwards aerodynamic force as the hovering helicopter, plus minor increases due to the vertical drag of the fuselage. The pilot transitioned the helicopter from hover to climb by pulling on the collective, increasing blade pitch and tilting the lift vector backwards (earth axes).

The vertical component of lift is equal to weight plus the downward vertical component of (blade drag + vertical fuselage drag). Lift is greater than its vertical component by a factor of 1/cos $\phi$.
So in this case (climb by increase in pitch):

Total upward vertical force is greater than weight by an amount of (D * sin($\alpha$) + vertical fuselage drag).
Lift is the only contributor to upward vertical force and is tilted back, so lift is greater than total vertical force by factor 1/cos($\alpha$).

Conclusion
Case 2. is considered multiple times on this site. Aerodynamic lift may be less than weight, depending on relevant angles and speeds. Thrust must always be higher than in steady horizontal flight by an amount of L * sin($\alpha$).
All cases have a higher upward vertical force than weight: a vertical aerodynamic drag component must be compensated for. Intuitively clear from this example.
@xxaviers answer is accepted. Many other answers are also correct for a steady state fixed wing climb due to aircraft axes tilt relative to gravity.

Good analysis! Indeed, lift is bigger than weight if thrust is not tilted (case 1) or is coupled with lift (case 3). In case 2, lift is smaller than weight if the vertical component of thrust is greater than the vertical component of drag. This is easily accomplished in almost all but the most extreme cases of downthrust because in a climb thrust is much bigger than drag in order to add potential energy to the plane. In a descent the reverse is true,

and again lift becomes smaller than weight (because now the small thrust points down and the bigger drag points up).

What exactly does "modulus" mean in this context?  Does it mean "magnitude"?  If so the answer might be clarified by making that switch.

Deleting comments; see "Addendum" at end of my answer https://aviation.stackexchange.com/a/56476/34686 for some thoughts on the "climbing by increasing airspeed" case, i.e. the  "climbing without allowing the aircraft to pitch up" case.

This answer could be improved by modifying or deleting this sentence: "All cases have a higher upward vertical force than weight."  The sentence only makes sense if you are adding up all the upward aerodynamic force components (including thrust), but not subtracting away the downward aerodynamic force components.

@quietflyer Consider a helicopter flying upwards, and the statement becomes intuitively clear - there is a downwards vertical component of the incoming airflow. Fixed wing aeroplanes are subject to the same physics laws as rotary wing aircraft...

The title of the question is different from the body of the question.
In the body of the question we read-

My question is purely about the sum of all vertical forces

Obviously for acceleration to be zero, net force must be zero, so net vertical aerodynamic force must be equal to weight.  (In this answer, we'll consider the thrust vector to be an aerodynamic force.)

in a steady climb, is the total upwards vertical force from all
sources (wing, tail, engines, fuselage) larger than, or equal to the
weight of the aircraft

In a steady climb, net vertical force must be zero, so net vertical aerodynamic force must be equal to weight.  However, this does not mean that the sum of all the upward vertical forces is equal to weight.  It is not, because one of the aerodynamic forces--the drag vector-- has a downward component in a climb.  Therefore the sum of all the upward vertical forces must be equal to weight plus the downward vertical component of the drag vector.
The title of the question, on the other hand, reads:

Does lift equal weight in a climb?

This is a distinctly different question-- and a more interesting question-- than a question about the net vertical force.
In the context of fixed-winged flight, Lift is defined to act perpendicular to the flight path through the airmass, and Drag is defined to act parallel to the flight path through the airmass.  For the purposes of the answer, we'll assume that Thrust acts parallel to the flight path through the airmass, although this clearly not always exactly true.  This simplifying assumption leads to the following vector diagram:
Powered climb at climb angles of 45 and 90 degrees:

In the vector diagrams above, "angle c" is the climb angle-- it is 45 degrees in the left-hand figure, and 90 degrees in the right-hand figure.
We can see that in a powered climb, Lift = Weight * cosine (climb angle), where the climb angle is measured relative to the airmass (an important distinction in the case of gliding flight-- an unpowered climb in a thermal updraft is still a descent in relation to the airmass!)
Clearly, Lift is less than Weight in a powered climb.  For example, if the climb angle is 45 degrees, Lift = .707 * Weight.  If the climb angle is 90 degrees, Lift must be zero.
The same is also true in a descent-- Lift = Weight * cosine (descent angle), so Lift is less than Weight.  This is explored in more detail in some of the links given at the end of this answer.
Note that we've taken the approach of combining the Thrust and Drag vectors into a single (Thrust-Drag) vector, and then we've arranged this vector into a closed vector triangle with Lift and Weight.  Whenever vectors can be arranged nose-to-tail into into a closed polygon-- a triangle in this case-- this shows that net force must be zero, meaning that acceleration is zero and velocity i constant.  For clarity we've also drawn the individual Thrust and Drag vectors outside the vector triangle.  These are redundant with the (Thrust-Drag) vector.
Varying the climb angle and/or the L/D ratio:

Note that for a given aircraft in a given configuration, any given angle-of-attack is associated with specific values for the lift coefficient, drag coefficient, and ratio of lift coefficient / drag coefficient.    Lift is proportional to lift coefficient * airspeed squared, and drag is proportional to drag coefficient * airspeed squared, so the ratio of the lift coefficient / drag coefficient is also the ratio of Lift / Drag.  So for a given aircraft in a given configuration, any given angle-of-attack is associated with specific ratio of Lift to Drag.
If the left-hand diagram above and the middle diagram above both represent the same aircraft in the same configuration, then the aircraft must be flying slightly slower in the middle diagram.  That's the only way the L and D values can both be slightly smaller, for the same L/D ratio.  Adding power to increase the climb angle, while holding the angle-of-attack constant, makes the airspeed decrease slightly.  However in the case illustrated here, the change in airspeed would be too tiny to ever notice in practice-- it would be equal to the square root of the change in the value of the magnitude of the lift vector or drag vector.
If all the diagrams represent the same aircraft in the same configuration of flaps etc, then the right-hand diagram (5:1 L/D ratio) would represent a lower angle-of-attack than the left-hand or middle diagrams (10:1 L/D ratio).  (We'll ignore the other possibility that the 5:1 case represents mushing flight very near stall, where drag is very high.)  A lower angle-of-attack means a lower lift coefficient, yet the size of the lift vector is the same, so the airspeed must be higher in the case illustrated in the right-hand diagram.  Therefore the climb rate is also higher.  In short, when we increase thrust to increase our climb rate, we also must reduce angle-of-attack, if for some reason we wish to keep our climb angle constant rather than allowing it to increase.
Powered climb at 45-degree climb angle at 8 different ratios of Lift to Drag:

Note that as we decrease our L/D ratio, it takes more and more thrust to maintain the same 45-degree climb angle.  In the case where the L/D ratio is 2/1, thrust must actually be greater than weight!  This is a bit counterintuitive, as we could obviously climb straight up with some small but non-zero airspeed if thrust were only slightly greater than weight.  However, that vertical climb would be conducted at a very low airspeed.  In the diagram above, if all cases represent the same aircraft in the same configuration, by constraining the climb angle to be constant, so that L also must remain constant, we're constraining the airspeed to get progressively higher and higher as we reduce the angle-of-attack, lift coefficient, and L/D ratio.  Hence the huge increase in drag, and thrust-required, as we reduce the angle-of-attack, lift coefficient, and L/D ratio.
As we explore climb angles closer and closer to 90 degrees, the L/D ratio has less and less influence on the thrust required.  A figure similar to the one above, but for a climb angle of 60 or 70 degrees, would show less increase  n thrust-required as we decrease the angle-of-attack, lift coefficient, and L/D ratio than we see at a climb angle of 45 degrees.  This also implies that we're forcing less of an increase in airspeed as we decrease the angle-of-attack, lift coefficient, and L/D ratio in such a case.  That makes sense-- as thrust carries more and more of the aircraft weight, the dynamics of the wing have less and less influence upon airspeed.  In the case of a truly vertical climb, the wing must be at the zero-lift angle-of-attack and the L/D ratio must be zero.  In such a case, of course, the drag force still varies with airspeed, and so the faster we want to fly straight up, the more thrust we need.
For the sake of clarity, this answer has focussed on some rather steep climb angles.  It's important to also keep in mind that for shallow climb angles (or descent) angles that are typical of general aviation light aircraft, the cosine of the climb angle is not much smaller than 1, and so Lift is nearly equal to Weight (specifically, Lift is only slightly less than Weight.)  Since Weight doesn't vary with climb or dive angle, we can conclude that for shallow climb or dive angles-- with no other accelerations going on (specifically, the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn)-- Lift is also nearly constant, regardless of whether the aircraft is climbing, descending, or neither.  This means if the climb or descent angle is shallow and the net G-load is one, the airspeed indicator can also be interpreted as an angle-of-attack gauge.  Why should this be so?  To keep lift nearly constant, it must be approximately true that the lift coefficient is varying in inverse proportion to the square of the airspeed. This establishes a nearly fixed relationship between airspeed and angle-of-attack, for shallow climb or descent angles and net G-loadings near one. If the airspeed is low, the lift coefficient and angle-of-attack must be high, and if the airspeed is high, the lift coefficient and angle-of-attack must be low, regardless of whether the aircraft is climbing at a shallow angle, descending at a shallow angle, or flying horizontally.  So the airspeed indicator is in essence an angle-of-attack gauge at shallow climb or descent angles. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- a given angle-of-attack will be associated with a lower airspeed than in horizontal flight, and a given airspeed will be associated with a lower angle-of-attack than in horizontal flight.  In the most extreme case where the aircraft is climbing straight up, lift must be zero, so the lift coefficient must be zero, and the angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.  Clearly the airspeed indicator cannot serve "double duty" as a guide to angle-of-attack in such a situation.
We've also assumed throughout this answer that the Thrust vector acts parallel to the flight path through the airmass.  Obviously, if this isn't true, then the equation lift = weight * cosine (climb angle) is also no longer true.  To take an extreme case, note that when the exhaust nozzles of a Harrier "jump jet" are pointed straight down, the wing is "unloaded"-- the plane can hover at zero airspeed with zero lift, supported entirely by thrust.  Conversely, during a glider winch launch, the towline pulls steeply downward on the glider.  This too can be viewed as a form of "vectored thrust"-- but now the load on the wing is increased, rather than decreased, so the wings must generate a lift force that is much greater than the aircraft's weight.  At any rate, it's best to thoroughly understand the simple case where the thrust vector acts parallel to the flight path, before going on to consider more exotic cases.
To see a vector diagram of the forces in climbing flight from an outside reference source, see the diagram below.  This diagram shows the same relationships as the other diagrams included in this answer, but the forces have not been arranged into a closed vector polygon, so it is less obvious that the net force is zero.

Above is a vector diagram showing the forces in a stabilized, linear, constant-airspeed climb -- from https://systemdesign.ch/wiki/L%C3%B6sung_zu_Steigflug
FS = thrust
FW = drag
FGp is the component of weight that acts parallel to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to (thrust - drag)
FGs is the component of weight that acts perpendicular to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to lift.
FA  = lift
FG = weight
Angle beta is the climb angle-- the angle between the flight path and the horizon.
Addendum:
Another answer to the present question addresses a  "climbing by increasing airspeed" case, which is also characterized by a completely fixed pitched attitude.  Perhaps a better description of this case is "entering a climb without pitching up at all".  This introduces a downthrust component relative to the flight path.  This downthrust is the root cause of the increase in the lift vector above and beyond the "standard" value of weight * cosine (climb angle).
The present answer assumes no downthrust or upthrust.  If downthrust or upthrust is present, the full equation for the magnitude of the lift vector is weight * cosine (climb angle) + thrust * sine (downthrust angle), where the downthrust angle is measured relative to the flight path, not the horizon.  Treat upthrust as negative downthrust.
When we start in level cruise with no downthrust, and then enter a climb without pitching up at all, the downthrust angle is now so large that lift is indeed greater than weight.  The climb angle and the downthrust angle are the same in this situation.  Note that as we increase thrust and begin climbing without allowing the aircraft to pitch up at all, the wing's angle-of-attack must decrease, which likely means that we are pushing forward on the control yoke or stick.  Naturally, this is not the normal way to conduct a climb!  See this ASE answer for more on the "climbing without allowing the aircraft to pitch up" case, in the context of an another analogous situation -- the "propeller train".
See these related answers to related questions:
"Are there any situations where having high lift but low lift to drag ratio would be beneficial?"
"What produces Thrust along the line of flight in a glider?"
"'Gravitational' power vs. engine power"
"Descending on a given glide slope (e.g. ILS) at a given airspeed— is the size of the lift vector different in headwind versus tailwind?"
"Are we changing the angle of attack by changing the pitch of an aircraft?"
"Is excess lift or excess power needed for a climb?"

Recent changes in this answer were motivated by my decision to delete another answer from another related question-- I noticed it had some content that fit well here.

Related: https://aviation.stackexchange.com/a/12720/34686

Re "A figure similar to the one above, but for a climb angle of 60 or 70 degrees, would show less increase n thrust-required as we decrease the angle-of-attack, lift coefficient, and L/D ratio than we see at a climb angle of 45 degrees."  -- should be "in" not "n".

@Therac -- no -- see the handwritten table at the bottom of the last of the hand-drawn diagrams.  I did mean 2/1.  Note that this is specifically for the case where the climb angle is constrained to be 45 degrees, so L is fixed, as is T-D.  See the T and D arrows at the top of that diagram, for the various cases of different L/D ratios, including the 2:1 ratio -- note that here, Thrust is greater than Weight.  Is that clear now?  Can you think of a way I could alter the text or diagrams to explain more clearly?  Or do you still think there's an error; if so could you say why?

Ah, OK, I can see it now. With the constraints of the case, and in the context of the diagram, that does work out this way.

No the lift will not equal the weight for an aircraft climbing (at constant velocity).
I cannot draw where I am at, so bear with me.
For an aircraft travelling at constant velocity, not undergoing acceleration, either vertically or horizontally, the lift generated by the wing will be less than the aircraft weight. You can see that the lift component will be less than the weight vector as you increase the climb angle. For example at a climb angle of 45 degrees, the lift component will equal square-root(2)/2 of the weight (or roughly 71% of the weight).
So how can the aircraft continue in a straight path upward? The engines provide thrust that applies a force equal to the difference in lift and weight. You can see this if you draw a force-balance diagram (which I will try to do later).

In flight dynamics (not aerodynamics), Lift is in the opposite direction along the stability Z axis. This is also where Lift Coefficient (CL) is calculated from.
It's not in the body axis, not in the earth axis, but in the stability axis. However you can use both wind Z axis and stability Z axis as they're the same for CL, but only stability X axis for CD.
Here's the definition by NASA TM 107601:

By DTIC ADA 189675:

Assuming the thrust is not deflected and is along the body X axis, then you can cast the thrust and weight onto the stability frame:

Forces on the stability Z axis: Lift + Thrust * sin(α) - Weight * cos(pitch-α)
Forces on the stability X axis: Thrust * cos(α) - Drag - Weight * sin(pitch-α)

Upon further consideration, I concur with the Answers concluding that, in a climb, Lift is less than Weight.
The applicable formula which appears in several engineering sources indicates that, in a climb: Lift = Weight X Sine(Climb Angle) and Thrust = Drag + Weight X Sine(Climb Angle) (See, e.g., this document.) Since the Sine is always less than 1 in a climb, Lift is always less than Weight in a climb. (Edit: Deleted reference to "vertical" Lift. The formula relates to total Lift since you want Lift = 0 in vertical climb.)
My initial Answer was that Lift is (of course) greater than Weight in a climb. This seemed to be "common sense" and was based on what I was taught during flight training. For example, according to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16]
It turns out that the FAA analysis is only partially correct. The real reason for increasing Lift is not to climb, but to change the flight path of the aircraft to match the desired Climb Angle. Once that is done, the aircraft is achieving most of the desired Climb Rate because the aircraft is now flying upwards. The FAA statement fails to mention that the main reason you need to add power is to maintain speed to offset the effects of gravity while you "climb the incline".

You say "the aircraft climbs because the force of lift acting vertically is greater than the force of gravity", but that would cause an acceleration, not a steady climb. You also say "You increase pitch to increase the angle of attack", but that is only true for a short moment because once you start climbing the angle of attack will reduce again (at the same pitch).

@Phil Crowther don't feel bad, this question is our local quicksand trap.  Around here, lift is defined as the force acting perpendicularly to the direction of flight.  So, as flight progresses to the vertical, the job of counter-acting gravity must be done more and more with thrust.  Excess lift is useless.  Force vector diagrams are very helpful.

Bianfable - (1) No, lift does not cause the aircraft to accelerate vertically. Interestingly, I was facing that same issue when attempting to simulate the vertical performance of a helicopter. I discovered that drag prevents positive thrust from generating perpetual acceleration. (2) No, the angle of attack does not reduce. it might try to do so, but you retrim the aircraft so that it continues to fly at the higher angle of attack. Robert- thanks for the kind words. I agree with what you say. I assume that the discussion is about shallow climbs, where lift is the dominant lifting force.

Regarding item 1, I should clarify that I mean parasitic drag operating vertically.  Just as with a helicopter climbing vertically, an aircraft that is climbing vertically will generate some parasitic drag in the vertical axis. It may not be much, but it is enough to offset the net acceleration and even help slow the vertical speed as the aircraft climbs into air that is less dense.  I believe that your observation would be correct if there was no air and we used a thruster to generate "lift". But we are not operating in a vacuum. I will post my helicopter computations in a separate topic.

@RobertDiGiovanni it's not "around here", it's the universally accepted definition: "Lift acts through the center of pressure of the object and is directed perpendicular to the flow direction." https://www.grc.nasa.gov/WWW/K-12/airplane/lift1.html

Fredrico, I believe that the technical definition is that lift acts "perpendicular to the chord of the airfoil". Most airplane designs include an angle of incidence that causes the chord to be angled back. This, along with changing angles of attack, mean that there is almost always a divergence between the direction of the real lift vector and a theoretical lift vector that is perpendicular to the direction of flight. However, because this difference is relatively small, most publications and most people tend to use a simpler definition.

Re "As the rate of climb increases to vertical, lift has less of a vertical impact and you have to rely more on thrust and momentum to climb" -- the reference to momentum doesn't pertain to a steady-state climb, which presumably was what the question was about.  Re "I work with vector-based flight simulations so I am working on an example which I hope will prove (or disprove) my point."-- did you look closely at https://aviation.stackexchange.com/a/56476/34686 ?

Re "Fredrico, I believe that the technical definition is that lift acts "perpendicular to the chord of the airfoil" -- that definition would totally invalidate the idea that the glide ratio of an unpowered aircraft is equal to the L/D ratio, which would invalidate a whole host of ASE questions and answers...

quiet flyer, I am not trying to invalidate the principles of aircraft aerodynamics. I am merely pointing out that there is more than one definition. For purposes of evaluating aircraft performance, we use the "perpendicular to the relative wind" definition and that makes perfect sense. However, if a wing works on the basis of differential pressure, it also makes perfect sense to conclude that the true lift vector operates "perpendicular to the wing chord".  Some commentators have pointed out that this "Total Lift" has a rearward component (which is generally small enough for us to ignore).

Re "My experience indicates that that there are cases where you can force an aircraft to climb using extra lift (as in the initial part of a climb from cruise speed) and that you might be able to sustain that with additional power. (Just like you can force an airplane to fly level with a high a-o-a and lots of power.)"  I think you are confusing an increase in lift coefficient with an increase in lift.  It might be helpful to read https://aviation.stackexchange.com/a/87040/34686 , which views the situation "from the other side"-- "is lift less in a descent than in horizontal flight"

And naturally, also https://aviation.stackexchange.com/a/56476/34686

@PhilCrowther flight will always be a combination of AOA and velocity (from thrust), so your second to last paragraph is on the right track and could be presented as a question.  Vy is all about climbing with the least amount of drag possible, using the excess Power (thrust x velocity) to its fullest.  "Learn as you answer" draws down votes, but it is good to get these concepts squared away.

The thing is, you could have calculated all that without knowing anything about the lift coefficient at all.  Say, a simple case where the pilot is manipulating the throttle as needed to hold speed constant.  Let's say we also assume we enter a climb via a constant radius of curvature of the flight path in the upward direction, and then at some point (via whatever pitch input is required) we instantly transition to a purely linear climbing path.  If we know the speed, and the radius of curvature during the climb entry, and the a/c weight, that's all we need to know to graph out the (ctd)

(Ctd) If we know the speed, and the radius of curvature during the climb entry, and the a/c weight, that's all we need to know to graph out the lift force generated throughout the entire maneuver. (Bearing in mind that the radius of curvature is influenced both by lift, and by the part of the gravity vector that acts perpendicular to the flight path.)  Lift will be greater than weight during the early part of the climb entry where the flight path is curving, but may become less than weight even before the flight path has stopped curving, and will certainly be < weight in the steady climb.

(Ctd) Of course a more representative climb entry would be different, like maybe hold power or thrust constant and ease back on stick to slowly let the airspeed decay till it stabilizes at the desired optimum value for max climb rate, etc...  but the point remains that if we know the trajectory, then we don't need to get into looking at the lift coefficient in order to calculate the lift force.  Though that method is valid as well, of course, as long as the calculations are done correctly (naturally).

The root problem with this answer is this-- "An equation that I found to compute cL required for climbing flight at the best climb speed (Vy) is: cL = sqrt(3 * pi * AR * e * cD0)."  I don't know where that comes from, but it must be based on some sort of estimate of the climb angle that will result.  It must be inherently built on the assumption that lift = weight* cosine climb angle.  So using this equation to calculate cL and then using that calculated cL to say something about the actual magnitude of the lift vector, is inherently circular logic, regardless of what answer you come out with.

But there's also something specifically wrong the suitability of the formula for this application, because you came out with the answer that the lift coefficient could be exactly the same when climbing at Vy as when flying horizontally at Vy, which isn't possible.  (Sorry for the wall of text but maybe it may help w/ improving the answer in some way?)

@PhilCrowther Indeed lift is less than weight if the fixed wing aircraft pitches up, the aircraft axis tilt relative to earth axes, and by definition part of the gravity vector is now in the aeroplane thrust/drag axis. But there are more cases to consider.

Re "The reason that I say Lift "should be" less than Weight is that it is possible that you could try to use more Lift to force the climb. But that would be the equivalent of slipping or skidding in a turn - a less efficient approach which wastes power."-- no, actually that would be the start of loop, or at least a "zoom" climb that gets continually steeper and steeper, with the airspeed continually decreasing--

Once more, my efforts to cover all the bases have gotten me in trouble. My calculations showed that, in a very limited range, you could fly inefficiently. But, as with the examples you mention, it could have been dynamically unstable, requiring constant correction. So I am going to delete that peripheral discussion and stick to what is covered by the equation.

@Koyovis - I almost hate to ask because, as noted above, my efforts to cover all the bases seem to keep getting me in trouble, but what are those other cases?

@PhilCrowther The confusing element that leads us into a vector jungle is the tilt of the aircraft axes relative to earth gravity axes. If we consider cases where the two axes systems are aligned, it is clearer to see. An F16 flying straight up at constant velocity, is thrust larger than weight? An aeroplane climbing by flying faster, is vertical upwards force larger than weight? A helicopter climbing in a hover, is total vertical force larger than weight? So yeah, if lift is defined ina colloquial way as the total vertical force, it is larger than weight. But lift is not defined that way...


text:
What is the mathematical relation between CLmax and Reynolds number?
I know that Reynolds number influences CLmax, but I am not able to find any formula with which I could plot a graph!

The Reynolds number is the ratio of inertial to viscous forces, so viscous effects become smaller with a growing Reynolds number. Viscous effects cause flow separation, so a higher Reynolds number helps to delay stall.
Details depend on the particular airfoil shape, so a general formula does not exist. However, there are many plots of how maximum lift varies over their Reynolds number for groups of similar airfoils.
Below you see Figure 18 from chapter 4 of S. Hoerner's "Fluid Dynamic Lift", giving the maximum lift coefficients of some airfoils of 12% thickness. Besides the Reynolds number, nose shape and camber also influence the maximum lift coefficient, so no simple formula is possible.

Note that the Reynolds number has a logarithmic scale and shows a linear dependency with lift coefficient above Re = 10⁷. So for large Reynolds numbers you might approximate the maximum lift as something like $$c_{L_{max}} = 0.3\cdot log(Re) - 0.5$$
but there is rather little experimental data to support or refine this approximation.

Can you also give me a relation between CLmax and mach number?

Yes. With the same caveats.